MHT CET · Physics · Mathematics in Physics
The resultant of two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) is \(\overrightarrow{\mathrm{R}}\). When the direction of \(\overrightarrow{\mathrm{Q}}\) is reversed, the
resultant is given by \(\overrightarrow{\mathrm{S}}\). Which one of the following is true for vectors \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{S}}\) ?
- A \(\mathrm{R}^{2}-\mathrm{S}^{2}=\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)\)
- B \(\mathrm{R}^{2}-\mathrm{S}^{2}=2(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}})\)
- C \(\mathrm{R}^{2}+\mathrm{S}^{2}=4(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}})\)
- D \(\mathrm{R}^{2}+\mathrm{S}^{2}=2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)\)
Answer & Solution
Correct Answer
(D) \(\mathrm{R}^{2}+\mathrm{S}^{2}=2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)\)
Step-by-step Solution
Detailed explanation
(B)
Given,
\(P+Q=R\)
After reversing direction of \(R\), we gel
\(\begin{array}{l}
r=-P-Q \\
S=-P-Q
\end{array}\)
So let angle between \(P\) and \(Q\) be \(Q\)
So resurtants.
\(\begin{array}{l}
R^{2}=p^{2}+Q^{2}+2 P Q \cos \theta...(1) \\
s^{2}=p^{2}+Q^{2}-2 P Q \cos \theta \quad...(2) \
\end{array}\)
Adthing equation (1) and (2)
\(R^{2}+S^{2}=2\left(P^{2}+Q^{2}\right)\)
So, the corvect answer is \(R^{2}+s^{2}=2\left(p^{2}+Q^{2}\right)\)
Given,
\(P+Q=R\)
After reversing direction of \(R\), we gel
\(\begin{array}{l}
r=-P-Q \\
S=-P-Q
\end{array}\)
So let angle between \(P\) and \(Q\) be \(Q\)
So resurtants.
\(\begin{array}{l}
R^{2}=p^{2}+Q^{2}+2 P Q \cos \theta...(1) \\
s^{2}=p^{2}+Q^{2}-2 P Q \cos \theta \quad...(2) \
\end{array}\)
Adthing equation (1) and (2)
\(R^{2}+S^{2}=2\left(P^{2}+Q^{2}\right)\)
So, the corvect answer is \(R^{2}+s^{2}=2\left(p^{2}+Q^{2}\right)\)
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